__Question 6.15__:
A
pump on the ground floor of a building can pump up water to fill a
tank of volume 30 m

^{3}in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

__Solution__:Volume of the tank,

*V*= 30 m

^{3}

*t*= 15 min = 15 × 60 = 900 s

Height of the tank,

*h*= 40 m

Efficiency of the pump, η = 30 %

Density of water,

*ρ*= 10

^{3}kg/m

^{3}

Mass of water,

*m*=

*ρ*

*V*= 30 × 10

^{3}kg

Output power can be obtained as:

P

_{0}= Work done / Time = mgh / t

= 30 X 10

^{3}X 9.8 X 40 / 900 = 13.067 X 10

^{3}W

For input power

*P*

_{i}

*,*, efficiency η, is given by the relation:

η = P

_{0}/ P

_{i}= 30%

P

_{i}= 13.067 X 100 X 10

^{3}/ 30

= 0.436 X 10

^{5}W = 43.6 kW

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